Problem 28 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{I}_{2}\) Equivalent weight of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is (a) \(\mathrm{M} / 3\) (b) \(\mathrm{M} / 6\) (c) \(\mathrm{M} / 2\) (d) none of these

Step-by-Step Solution

Answer

The equivalent weight of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is \(\mathrm{M} / 6\).

Step 1: Understand the Reaction

The given reaction is a redox reaction involving redox changes of chromium from Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\) (reduction) and iodine from \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\) (oxidation). We need to find the equivalent weight of the dichromate \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\).

Step 2: Determine the Change in Oxidation Number

In \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), chromium starts with an oxidation number of +6 and is reduced to +3 in \(\mathrm{Cr}^{3+}\). The change in oxidation state for one \(\mathrm{Cr}\) atom is \(+6 - (+3) = 3\). Since there are two \(\mathrm{Cr}\) atoms, the total change is \(2 \times 3 = 6\).

Step 3: Calculate the Molar Mass of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)

The molar mass of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is composed of 2 Cr atoms and 7 O atoms. Using chromium's atomic mass of 52 and oxygen's atomic mass of 16, the molar mass is calculated as follows: \[ (2 \times 52) + (7 \times 16) = 104 + 112 = 216 \text{ g/mol} \].

Step 4: Calculate the Equivalent Weight of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)

The equivalent weight is given by the formula: \(\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Change in oxidation number}}\).Using the molar mass from Step 3 and the change in oxidation number from Step 2, the equivalent weight is: \[ \frac{216}{6} = 36 \text{ g/equiv} \].

Step 5: Choose the Correct Answer

The given options are expressed in terms of \(M\), which typically represents the molar mass in questions involving equivalent weight. Therefore, the equivalent weight \(\mathrm{M}/6\) matches our calculated equivalent weight of 36 g/equiv, since \(M = 216\). Hence, option (b) \(\mathrm{M} / 6\) is the correct answer.

Key Concepts

Redox Reaction

A redox reaction, short for reduction-oxidation reaction, is a chemical process in which oxidation and reduction occur simultaneously. In simple terms, oxidation involves the loss of electrons, while reduction involves the gain of electrons. These processes happen together because when one substance loses electrons, another must gain them to maintain balance.

In the given chemical equation: \[\text{Cr}_2 \text{O}_7^{2-} + 6 \text{I}^- + 14 \text{H}^+ \longrightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2 \text{O} + 3 \text{I}_2\]chromium in \(\text{Cr}_2 \text{O}_7^{2-}\) is reduced to \(\text{Cr}^{3+}\), indicating it has gained electrons (reduction). In contrast, the iodide ions \(\text{I}^-\) are oxidized to form iodine \(\text{I}_2\), showing they have lost electrons (oxidation).

Redox reactions are key in various fields such as energy production, metabolism, and industrial chemical processes. They illustrate the interplay of electron transfer, powering the mechanisms behind batteries, respiration, and other essential processes in life and technology.

Oxidation State

An oxidation state, also known as an oxidation number, represents the hypothetical charge an atom would have if all bonds to atoms of different elements were completely ionic. It provides insight into the electron transfer that occurs during a chemical reaction.

To determine the oxidation state, one follows some rules:

• The oxidation number of a free element is always zero.
• The oxidation number of a monoatomic ion equals the ion's charge.
• Oxygen generally has an oxidation number of -2 in compounds, while hydrogen is usually +1.
• The sum of oxidation numbers in a neutral compound is zero, and for a polyatomic ion, it is equal to the charge of the ion.

In this exercise, chromium in \(\text{Cr}_2 \text{O}_7^{2-}\) starts with an oxidation state of +6. Each chromium atom undergoes a change to +3 when forming \(\text{Cr}^{3+}\). This helps us calculate the equivalent weight by revealing the total change in the oxidation number.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Calculating molar mass is crucial for understanding the proportion of elements in chemical compounds.

To find the molar mass, follow these steps:

• Identify the atomic masses of each element (found on the periodic table).
• Multiply the atomic mass of each element by the number of atoms of that element in the compound.
• Sum these values to find the total molar mass.

In our example, the molar mass of\(\text{Cr}_2 \text{O}_7^{2-}\) was evaluated by calculating:\[(2 \times 52) + (7 \times 16) = 104 + 112 = 216 \text{ g/mol}.\]This molar mass allows us to determine the equivalent weight, a critical factor when dealing with redox reactions and stoichiometry problems. The molar mass is an essential value, especially in calculating how much of each substance is needed or produced in a chemical reaction, guiding correct proportions for experimental and industrial processes.